# If a sample originally had 120 atoms of carbon-14, how many atoms will remain after 16,110 years?

Mar 12, 2016

We know that carbon-14 ($\text{^14 "C}$) is an unstable isotope (it's not $\text{^12 "C}$, and we say some will become something else after $16110$ years, because it's radioactive).

So, $\text{^14 "C}$ undergoes radioactive decay, a generally zero-order process that follows a half-life decay mechanism.

The half-life with respect to the rate constant would be found to be:

${t}_{\text{1/2}} = \frac{\ln 2}{k}$

where the point of showing the equation is that it does not depend on the concentration of $\text{^14 "C}$, which is what a zero-order process is.

Logically, what happens in a half-life decay is that half the sample becomes something else, so of course, when we focus on the original sample, we would see that half is left when the process has occurred once.

After each half-life passes by, we lose half of the remaining sample, so you can see that we can simply halve the concentration $n$ number of times and we get the concentration after $n$ half-lives.

It therefore makes sense that after $n$ half-lives, we get $\frac{1}{2} \cdot \frac{1}{2} \cdot \cdot \cdot \cdot \frac{1}{2} = {\left(\frac{1}{2}\right)}^{n} = \frac{1}{{2}^{n}}$ times the original concentration. Thus, we would construct this equation to determine the concentration after $n$ half-lives:

$\textcolor{b l u e}{{\left[\text{^14 "C"]_f = 1/(2^n)[""^14 "C}\right]}_{i}}$

Using the half-life of carbon-14, which is known to be about $5730$ years, we get:

= 1/(2^("16110/5730"))[""^14 "C"]_i

$\frac{16110}{5730} \approx 3$, so we should expect approximately $\text{1/8}$ of the original concentration remains, or about $15$ atoms.

We would actually see that this gives the final concentration of $\text{^14 "C}$ to be a bit above $\text{1/8}$ of the original concentration. Specifically, we would get about $17.093$, or about $\setminus m a t h b f \left(17\right)$ atoms remaining.